[leetcode daily question] a single element in an ordered array

A single element in an ordered array
difficulty ： secondary

The time complexity required for the topic is log n, It's not hard to think of solving it through binary search , The observation topic is easy to come to , Only once Subscripts are even .
The left bound of binary search is 0, The right boundary is the maximum subscript of the array . Take the average value of the left and right boundaries each time mid As a subscript to be judged , according to mid The parity of determines the comparison with the adjacent elements on the left or right ：

• If mid It's even , Then compare nums[mid] and nums[mid+1] Whether it is equal or not ;
• If mid Is odd , Then compare nums[mid−1] and nums[mid] Whether it is equal or not .

If the result of the above comparison of adjacent elements is equal , be mid<x, Adjust the left border , otherwise mid≥x, Adjust the right border . After adjusting the boundary, continue the binary search , Until the subscript is determined x Value .
Get the subscript x After the value of ,nums[x] That is, the element that appears only once .
The code is as follows ：

	 public int singleNonDuplicate(int[] nums) {
    
        int left = 0;
        int right = nums.length-1;
        if(nums.length==1||nums[0]!=nums[1]){
    
            return nums[0];
        }
        while (left<right){
    
            int mid = (left+right)/2;
            if (mid%2==0){
    
                if (nums[mid]==nums[mid+1]){
    
                    left=mid+1;
                }else if (nums[mid]==nums[mid-1]){
    
                    right = mid-1;
                }else{
    
                    return nums[mid];
                }
            }else{
    
                if (nums[mid]==nums[mid-1]){
    
                    left=mid+1;
                }else if (nums[mid]==nums[mid+1]){
    
                    right = mid-1;
                }else{
    
                    return nums[mid];
                }
            }
        }
        return nums[left];
    }

Execution results ： success